What is the DFT of DFT of discrete signal











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What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?










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    What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?










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      up vote
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      down vote

      favorite











      What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?










      share|improve this question













      What is the discrete fourier transform of the discrete fourier transform of any discrete time signal. Is result same signal? How?







      discrete-signals fourier-transform dft






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      asked 8 hours ago









      Mert Ege

      133




      133






















          2 Answers
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          let



          $$begin{align}
          X[k] &= mathcal{DFT} Big{ x[n] Big} \
          &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
          end{align} $$



          and



          $$ y[n] triangleq X[n] $$



          (note the substitution of $n$ in for $k$.) then



          $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



          then, if the DFT is defined the most common way (as above):



          $$ Y[n] = N cdot x[-n] $$



          where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






          share|improve this answer




























            up vote
            1
            down vote













            Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



            Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






            share|improve this answer























            • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
              – robert bristow-johnson
              7 hours ago











            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            4
            down vote













            let



            $$begin{align}
            X[k] &= mathcal{DFT} Big{ x[n] Big} \
            &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
            end{align} $$



            and



            $$ y[n] triangleq X[n] $$



            (note the substitution of $n$ in for $k$.) then



            $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



            then, if the DFT is defined the most common way (as above):



            $$ Y[n] = N cdot x[-n] $$



            where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






            share|improve this answer

























              up vote
              4
              down vote













              let



              $$begin{align}
              X[k] &= mathcal{DFT} Big{ x[n] Big} \
              &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
              end{align} $$



              and



              $$ y[n] triangleq X[n] $$



              (note the substitution of $n$ in for $k$.) then



              $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



              then, if the DFT is defined the most common way (as above):



              $$ Y[n] = N cdot x[-n] $$



              where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






              share|improve this answer























                up vote
                4
                down vote










                up vote
                4
                down vote









                let



                $$begin{align}
                X[k] &= mathcal{DFT} Big{ x[n] Big} \
                &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
                end{align} $$



                and



                $$ y[n] triangleq X[n] $$



                (note the substitution of $n$ in for $k$.) then



                $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



                then, if the DFT is defined the most common way (as above):



                $$ Y[n] = N cdot x[-n] $$



                where periodicity is implied: $x[n+N]=x[n]$ for all $n$.






                share|improve this answer












                let



                $$begin{align}
                X[k] &= mathcal{DFT} Big{ x[n] Big} \
                &triangleq sumlimits_{n=0}^{N-1} x[n] , e^{-j2pi nk/N}
                end{align} $$



                and



                $$ y[n] triangleq X[n] $$



                (note the substitution of $n$ in for $k$.) then



                $$ Y[k] = mathcal{DFT} Big{ y[n] Big} $$



                then, if the DFT is defined the most common way (as above):



                $$ Y[n] = N cdot x[-n] $$



                where periodicity is implied: $x[n+N]=x[n]$ for all $n$.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered 7 hours ago









                robert bristow-johnson

                10.4k31448




                10.4k31448






















                    up vote
                    1
                    down vote













                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






                    share|improve this answer























                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago















                    up vote
                    1
                    down vote













                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






                    share|improve this answer























                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago













                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).






                    share|improve this answer














                    Depends on how the DFT implementation or equation is scaled and indexed. The result of dft(dft(x)) is to circularly reverse the array x (of length N) around its first element, possibly with a scale factor of N, 1/N, or 1/sqrt(N).



                    Computationally, there may also be added numerical or quantization noise (for instance, to the imaginary components if they were originally all zero for strictly real input).







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited 7 hours ago

























                    answered 7 hours ago









                    hotpaw2

                    25.5k53472




                    25.5k53472












                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago


















                    • shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                      – robert bristow-johnson
                      7 hours ago
















                    shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                    – robert bristow-johnson
                    7 hours ago




                    shouldn't matter how it's indexed (as best as i can understand it). you're certainly correct about the scaling, hot.
                    – robert bristow-johnson
                    7 hours ago


















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